The capacitance of capacitor will be:
A) \[\frac{{{\varepsilon }_{0}}A({{K}_{1}}+{{K}_{2}})}{2d}\]
B) \[\frac{{{\varepsilon }_{0}}A}{2d}\left( \frac{{{K}_{1}}+{{K}_{2}}}{{{K}_{1}}{{K}_{2}}} \right)\]
C) \[\frac{{{\varepsilon }_{0}}}{d}\left( \frac{{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}} \right)\]
D) \[\frac{{{\varepsilon }_{0}}A}{d}\left( \frac{{{K}_{1}}+{{K}_{2}}}{{{K}_{1}}{{K}_{2}}} \right)\]
Correct Answer: A
Solution :
Key Idea: In each capacitor, the area of the plate will be \[\frac{A}{2}.\] In the given combination, the arrangement is equivalent to two capacitors connected in parallel. Also in each capacitor the area of the plate will be \[\frac{A}{2}.\] Therefore, equivalent capacitance \[C={{C}_{1}}+{{C}_{2}}\] \[=\frac{{{K}_{1}}{{\varepsilon }_{0}}A/2}{d}+\frac{{{K}_{2}}{{\varepsilon }_{0}}A/2}{d}\] \[=\frac{{{\varepsilon }_{0}}A}{2d}({{K}_{1}}+{{K}_{2}})\]
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